Since we sample two cameras, the random variable X can only have three possible values:
0 - both cameras work correctly
1 - only one defective camera
2 - both cameras are defective.
To fully define the random variable, we need to compute the proability for each of the possible values.
X follows a Bernoulli distribution, where we want to count the number of "successes" (finding a defective camera) on a sample of 6 elements. Since 4 out of the 6 cameras are defective, the probability of finding a defective camera is 4/6=2/3.
In general, we have
[tex] P(X = k) = \binom{2}{k} \left(\cfrac{2}{3}\right)^k\left(\cfrac{1}{3}\right)^{2-k} [/tex]
So, the probabilities are
[tex] P(X = 0) = \binom{2}{0} \left(\cfrac{2}{3}\right)^0\left(\cfrac{1}{3}\right)^{2} = \cfrac{1}{9} = 0.\overline{1} \approx 0.11 [/tex]
[tex] P(X = 1) = \binom{2}{1} \left(\cfrac{2}{3}\right)\left(\cfrac{1}{3}\right) = \cfrac{4}{9} = 0.\overline{4} \approx 0.44 [/tex]
[tex] P(X = 2) = \binom{2}{2} \left(\cfrac{2}{3}\right)^2\left(\cfrac{1}{3}\right)^{0} = \cfrac{4}{9} = 0.\overline{4} \approx 0.44 [/tex]
The random variable, x which defines the number of defective cameras in the selected sample is ; x = 0, 1, 2
The number of defective cameras in the box = 2
Since two selection will be Made ;
The random variable, x is the sample space for the possible Number of defective cameras that can be selected ;
The possibilities are :
Therefore, the definition of the random variable, x = 0, 1, 2
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